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NIOS Class 12 Biology Chapter 23 Molecular Inheritance and Gene Expression
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Molecular Inheritance and Gene Expression
Chapter: 23
MODULE – III: REPRODUCTION AND HEREDITY
NIOS TEXTBOOK QUESTIONS ANSWERS
INTEXT QUESTIONS 23.1
1. Expand the abbreviation DNA.
Ans: DNA: Deoxyribonucleic Acid.
2. Name the scientists who confirmed that DNA was the genetic material in bacterial transformation.
Ans: A very, McLeod and McCarty.
3. Name the sugar and the nitrogenous bases found in DNA.
Ans: Deoxyribose:
(i) Adenine.
(ii) Guanine.
(iii) Cytosine. and
(iv) Thymine.
INTEXT QUESTIONS 23.2
1. In which direction does DNA polymerase proceed to catalyse DNA replication 5′ to 3′ or 3′ to 5′?
Ans: In 5′-3′ direction.
2. What is a primer ___________ a DNA molecule or an RNA molecule?
Ans: RNA molecule.
3. Name the four enzymes needed for DNA replication.
Ans: (i) Helicase.
(ii) DNA polymerase.
(iii) DNA ligase. and
(iv) Topoisomerase.
4. Which enzyme joins the okazaki pieces to form a complete DNA strand?
Ans: DNA ligase.
INTEXT QUESTIONS 23.3
1. What is central dogma in molecular Biology?
Ans: It refers to “the transfer of genetic information from genes to the site of protein synthesis.”
2. Which molecule is synthesised at transcription?
Ans. Cistronic DNA.
3. What is a codon? What is meant by ‘code is degenerate’?
Ans: It is the sequence of 3 bases in the genes.
Codon degeneracy means that the genetic code is degenerate. It means that there is more than one codon that specifies the single amino acid. The phenomenon in which several of the amino acids, each coded by more than one codon is called codon degeneracy.
4. Where in the cell does translation occur?
Ans: Nucleus.
5. Name the three types of RNA that participate in protein synthesis.
Ans: (i) mRNA.
(ii) t-RNA. and
(iii) r-RNA.
INTEXT QUESTIONS 23.4
1. Name the components of an operon.
Ans: Operator, regulator and structural genes.
2. What is mutation? When is a mutation called a transition mutation?
Ans: Mutation: A heritable change in the structure, con-tent and organisation of genetic material. When in a DNA sequence a purine is replaced by purine and pyrimidine is replaced by pyrimidine.
3. Why is “silent mutation” called so?
Ans: A silent mutation does not being about a change in the sequence of nucleotides.
4. What are mutagens?
Ans: These are agents which cause mutation in genetic material.
5. Name a chemical which causes mutation in the heredity material.
Ans: X-rays.
TERMINAL EXERCISES
1. How did Hershey and Chase prove that DNA is the hereditary material?
Ans: Hershey and Chase (1952) performed an experiment with two different preparations of bacteriophage (T₂ phase). In one preparation,they used radioactive protein and in the DNA was made radioactive. These two T₂ phase preparations were made to infect a culture of E.coli. Soon after the E.coli cells were gently agitated in a blender to loosen the ordering phase particles and the culture was centrifugal. The heavier infected bacterial cells settled at bottom. The lighter virus particles and other viral components had not entered the bacterium. Thus the protein coat (for which radioactive of sulphur S¹⁵ was used) does not enter during the infection.
In another experiment when T₂ phage containing DNA (for which radioactive of phosphorus-P³²) was used to infect E.coli, the heavier bacterial pellet was radioactive. Hershey and Chase found that a radioactive isotope of phosphorus gets incorporated in DNA rather than into protein. Thus the experiment indicated that during the infection with virus, it is the DNA that actually entered in the bacteria. It suggests that DNA and not the protein, is the genetic material.
| S³⁵ Tracer for proteinT2 – Bacteriophage used in inject E.coli. (Experiment) ________ T₂ Bacteriophage ________ E.coli on S³⁵ medium Bacteriophage injected lysis DNA, radiactive T₂ phage S35 protein DNA Non-isotopic New E.coliDNA-Non Radioactive | p³² for DNA E.coli P³¹ medium lysis Used to inject Normal E. coli New normal E. colilysisDNA, Radioactive |
Fig. 23.13. Diagrammatic representation.
2. Explain:
(i) Transduction.
Ans: Transduction: It refers “to transfer of DNA from one bacterial cell into another bacterium through the agency of a virus.”
(ii) Lysogeny.
Ans: Lysogeny: One of two outcomes of the infections of the host cell by a temperate phage. One outcome is that the phage genome becomes repressed and the phage DNA replicates as part of the host DNA forming a Lysogen, infrequently the Lysogen may become induced and the host cell may burst releasing a number of phage particles. The other outcome is the Lytic cycle that produces progeny phase particles.
3. Describe the Watson and Crick Model of DNA.
Ans: Watson and Crick Model of DNA: According to Watson and Crick (1953) the DNA molecule consisted of two strand twisted around each other in the form of a helix. Each strand being made of polynucleotides. Each polynucleotide consisting of many nucleotides which remain united with its complimentary chain with the help of bases. Adenine always unites with thymine and cytosine with guanine. It means that one polynucleotide chain of DNA molecule is complementary to the other.
Fig. 23.14. Watson and Crick model of DNA (Nucleic acid).
The distance between two chains of the helix is about 20 Å and helix over every 34Å. Each turn of the chain consists of about 10 nucleotides.
3′-5′ chain: The two strands of DNA are in antiparallel direction or opposite direction. It states that the carbon atom at 5′ position in the sugar is in one direction in one strand and in opposite direction in the other strand. So the two strands are antiparallel (their 5′-3′ direction remains opposite).
Fig. 23.15. Four nucleotides contained in DNA molecule.
4. Explain how replication takes place.
Ans: DNA is “master molecule of body”. Messelson and Stahl showed that “DNA replication is semiconservative.” E.coli needs a set of enzymes for it-DNA dependent DNA polymerase. In E.coli replication is completed in 38 minutes (average 2000 bp per second). Many other enzymes are also needed. A particular form, where the intertwined DNA start separating is known as the origin of replication. Helicases are the enzymes which unwind DNA helix. Topoisomerases are the enzymes which can break and reseal one strand of DNA. Primer is a short stretch of RNA formed on DNA templates. Primase is the enzyme which polymerises the RNA building blocks (AUGC) into the primer.
Mechanism of Replication of DNA: It is complicated and several enzymes take part in it. It was suggested by Watson and Crick. It was semi-conservative. The enzyme DNA poly-merase helps in adding the building blocks to the primer in a sequence as influenced by the template. The small fragments of DNA formed on lagging strand in 5′ – 3′ direction starting from a DNA primer during replication of DNA are called okazaki fragments. (These fragments are joined together by the enzyme DNA ligase). The strand that supports continuous DNA synthesis is leading strand which is replicated in short stretches is lagging strand.
5. Write note on Central Dogma.
Ans: Central Dogma: It is defined “as the transfer of information from genes to site of protein synthesis”. It operates in sequence: information flows from DNA to specific protein via RNA as:
To synthesis protein information coded in DNA is copied as a complementary messenger RNA; it is transciption. Messenger RNA carrying information moves out of nucleus into cytoplasm → attaches to ribosomes for translate action of information in form of a protein translation. But in retroviruses, genetic material is RNA; hence, during protein synthesis it is first ‘transcribed into a DNA molecule in presence of Reverse transcriptase path of central dogma is now:
6. State the properties of the genetic code.
Ans: The properties of Genetic Code:
(i) The genetic code is universal: In the world, all the organisms whether plant, animal or even a virus have 64 codons to represent genetic message.
(ii) The genetic code is non-overlapping: Same nucleotide cannot function as component of the next code in its succession. The genetic code is linear.
(iii) The code is triplet (Sixty one codon code for amino acids).
7. Explain transcription in Eukaryotes and processing of hnRNA.
Ans: Transcription In Eukaryotes: The flow of genetic information from cistronic DNA to mRNA is termed transcription. It occurs in the steps given below:
(a) Cistronic DNA (which carries the informa- tion for protein to be synthesised) unwinds in the presence of helicase and topoisomerase enzyme.
(b) RNA polymerase (enzyme) begins to catalyse synthesis of mRNA signalled by a protein the sigma factor.
(c) mRNA is synthesised. It complementary to cistronic DNA and a Rho factor signals RNA polymerase to complete transcription process.
(d) Strand of DNA that has code for transcription of specific protein is known as sense strand of DNA opposed to antisense strand, that which is not transcribed.
Fig. 23.17. Transcription in prokaryotes.
(e) In eukaryotes, mRNA is synthesised within nucleus after cistronic DNA unwinds to expose sense strand for transcription of a molecule of RNA termed as mRNA, is synthesised of RNA polymerase.
The Processing of mRNA: mRNA is large since the eukaryotic genes have coding sequences called exons of non-coding sequences called introns (I) in between exons. Both there introns and exons (E) are transcribed in mRNA. During processing of mRNA the introns are removed but the exons join to make mRNA.
(a) A nucleoside known as the methyl guanosine comes and attaches at 5′ end of mRNA. This is called capping.
(b) A small piece of RNA having only nucleotides containing base Adenine is attached at 3′ end. It is termed as the poly A tail.
(c) mRNA with cap and tail moves out of pores in the nuclear membrane.
(d) Process of formation of functional mRNA from hnRNA is called RNA processing. See Fig. 22.18.
Fig. 23.18. Schematic drawing showing transcription and processing of hnRNA in eukaryotes.
8. What do you mean by regulation of genes?
Ans: Regulation of Genes: All cells possess genes but only those function that needed to be active in multi celled organisms.
The expression of genes is “regulated by switching on and switching off genes whenever needed.”
9. Explain how the lac operon gets switched on in the presence of lactose in E. coli.
Ans: Lac-operon in E. coli: According to F. Jacob and J. Monod, in E. coli, the metabolic pathways are regulated as a unit. When the sugar lactose is added to the culture of E. coli three enzymes-ß galactosidase, permease and transacetylase are produced. The genes for these three enzymes are linked and are known as structural genes.
These structural genes have the information to code for the amino acid sequence and thus directly decide the structure and function of the individual protein of the pathway.
These three genes, coding for three enzymes are regulated as unit by a single switch known as operator. This entire unit is known as operon. The operator gene is controlled by a regulator gene through regulator protein.
The expression of genes depends upon on or off of the switch. When switch is on the three genes are transcribed by RNA polymerase into a single stretch of mRNA covering all the three genes: mRNA translates into proteins.
DNA → mRNA → Proteins
When lactose is added a few molecules of it get into the cell by the action of the enzyme permease. These few molecules are then converted into an active form of lactose, which binds to the repressor protein. Thus the repressor can not bind to the operator switch.
But when lactose is not added, the repressor protein bind with the operator and the genetic system remain in switch off position. Thus there will be no transcriptions and hence no translation; thereby no mRNA synthesis and no protein formation occurs.
Fig. 23.19. Lactose operon consisting of various genes for lactose digestion.
10. Name three levels at which regulation takes place in a eukaryotic cell.
Ans: Gene regulation in the eukaryotic cell is complex. It occurs in several ways. The gene expression can be regulated as following:
(a) at the level of transcription.
(b) at the level of processing of hnRNA into m-RNA.
Fig. 23.20. Levels of gene control in Eukaryotes.
(c) at translation or even.
(d) at post translation.
11. Write notes on:
(i) Types of mutations.
Ans: Types of Mutations: Mutation is “a heritable change in the structure of gene or chromosome or in number of chromosomes.” It occurs due to various mutagens like UV rays, X-rays and certain chemicals.
Mutation in a gene is called gene/point mutation. Chromosomal mutation occurs in the structure and number of chromosomes. Euploidy refers to change in number of chromosomes e.g., 4N, 6N etc. Aneuploidy is the loss/addition of one or more chromosomes in a set. Change in structure of chromosomes is called chromosomal aberrations.
It may be due to deletion, duplication, inversion or translocation.
Gene mutation may be of following types:
(i) Transition.
(ii) Frameshift.
(iii) Transversion.
(iv) Mis-sense.
(v) Non-sense.
(vi) Silent etc.
(ii) Okazaki fragments.
Ans: Okazaki Fragments: The small fragments of DNA formed on lagging strand in 5′-3′ direction starting from a DNA primer during replication of DNA are called okazaki fragments. (These fragments are joined together by the enzyme DNA ligase).
(iii) Chain termination during translation.
Ans: Termination of Chain: There are three codons on the 3rd end of mRNA for which no tRNA is available for interaction. When these codons are exposed, no amino acid would be added since a nonsense codon is not usually recognized by tRNA and resulting in stop in protein synthesis. These codons are called nonsense codons or terminating codons or stop signals. They are UAA (Ochre), UAG (Amber) and UGA (Opal).
